Question: Is ${337726}$ divisible by $3$ ?
Solution: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {337726}= &&{3}\cdot100000+ \\&&{3}\cdot10000+ \\&&{7}\cdot1000+ \\&&{7}\cdot100+ \\&&{2}\cdot10+ \\&&{6}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {337726}= &&{3}(99999+1)+ \\&&{3}(9999+1)+ \\&&{7}(999+1)+ \\&&{7}(99+1)+ \\&&{2}(9+1)+ \\&&{6} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {337726}= &&\gray{3\cdot99999}+ \\&&\gray{3\cdot9999}+ \\&&\gray{7\cdot999}+ \\&&\gray{7\cdot99}+ \\&&\gray{2\cdot9}+ \\&& {3}+{3}+{7}+{7}+{2}+{6} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${337726}$ is divisible by $3$ if ${ 3}+{3}+{7}+{7}+{2}+{6}$ is divisible by $3$ Add the digits of ${337726}$ $ {3}+{3}+{7}+{7}+{2}+{6} = {28} $ If ${28}$ is divisible by $3$ , then ${337726}$ must also be divisible by $3$ ${28}$ is not divisible by $3$, therefore ${337726}$ must not be divisible by $3$.